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This is what I got. $\dfrac{1}{6} \cdot \dfrac{1}{6} = 2.78\% \cdot 24 = 66.72\%$

I believe that since it is a six sided dice, since you roll both of them simultaneously it would be $\dfrac{1}{6} \cdot \dfrac{1}{6}$.

So since they are rolling them $24$ times, I would just multiply it by $24$, so $2.78\% * 24$ would be $66.72\%$, which would mean I have a $67.7\%$ chance of rolling a double six.

Do you think this is correct? Am i doing this correctly?

Mhenni Benghorbal
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Ghozt
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    That's not the right method; imagine rolling the dice 1000 times - what would you conclude the chance was now? – not all wrong May 21 '13 at 09:39
  • If that were the probability of getting a double six in 24 trials, then the probability of getting a double six in 72 trials would be $2$. But no probability can be more than $1$. – Michael Hardy May 21 '13 at 13:10
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    In fact $\frac16\cdot\frac16\cdot24$ is EXACTLY $2/3$, as can be seen by canceling. That rounds to $66.67%$, not $66.72%$. Your mistake in arithmetic comes from rounding too early. – Michael Hardy May 21 '13 at 13:12

1 Answers1

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We calculate the probability of rolling at least one double-six in $24$ rolls of two dice. The probability we roll a double-six is, as you point out, $\frac{1}{36}$.

So, on any roll, the probability of not getting a double six is $\frac{35}{36}$.

The probability of "failure" $24$ times in a row is therefore $\left(\frac{35}{36}\right)^{24}$.

So the probability of at least one "success" in $24$ rolls is $1-\left(\frac{35}{36}\right)^{24}$.

Remark: If we interpret the question as asking for the probability of exactly one double-six, the answer is different, and more complicated. In that case, we want the probability of $1$ success and $23$ failures. This probability is $\binom{24}{1}\left(\frac{1}{36}\right)\left(\frac{35}{36}\right)^{23}$.

However, the usual interpretation is "at least one six."

This problem goes back to the the seventeenth century beginnings of probability theory. The Chevalier de Montmort was curious about what was more likely, at least one six in $4$ rolls of a die, or at least one double-six in $24$ rolls of two dice. Each has probability about $50\%$ and the expected number of successes is the same. At the time, expectation was the primary notion.

André Nicolas
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  • Nicolas, would the probability of rolling one six in 4 tosses of a die be 1-(5/6)^4? – Ghozt May 21 '13 at 09:59
  • That is correct if you mean the probability of rolling at least one six. – André Nicolas May 21 '13 at 10:05
  • You are welcome. It is clear that you understand the idea. Fairly often, it is easier to find the probability that $A$ happens by first finding the probability $A$ doesn't happen. – André Nicolas May 21 '13 at 10:11
  • Wouldnt nicolas' answer be for tossing the die 48 times? Since 1/36 is the probability of getting a pair of 6 in tossing the die twice, shouldnt it be (35/36)^12 ? –  Oct 06 '13 at 08:16
  • The standard interpretation of the "$24$ rolls" is that two dice are rolled each time. True, that is $48$ individual dice rolls. However, we are interested in the event "double $6$" so it is how the $48$ rolls combine in $24$ pairs that matters. – André Nicolas Oct 06 '13 at 16:13